Question -
Answer -
Given:
In a ∆ABC, a = √2, b = √3 and c = √5
By using the formulas,
We know, cos A = (b2 + c2 –a2)/2bc
By substituting the values we get,
= [(√3)2 + (√5)2 –(√2)2] / [2 × √3 × √5]
= 3/√15
We know, Area of ∆ABC = 1/2 bc sin A
To find sin A:
Sin A = √(1 – cos2 A) [by usingtrigonometric identity]
= √(1 – (3/√15)2)
= √(1- (9/15))
= √(6/15)
Now,
Area of ∆ABC = 1/2 bc sin A
= 1/2 × √3 × √5 × √(6/15)
= 1/2 √6 sq. units
Hence proved.